""" Problem Statement (Digit Fifth Power ): https://projecteuler.net/problem=30
Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
1634 = 1^4 + 6^4 + 3^4 + 4^4
8208 = 8^4 + 2^4 + 0^4 + 8^4
9474 = 9^4 + 4^4 + 7^4 + 4^4
As 1 = 1^4 is not a sum it is not included.
The sum of these numbers is 1634 + 8208 + 9474 = 19316.
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
(9^5)=59,049
59049*7=4,13,343 (which is only 6 digit number )
So, number greater than 9,99,999 are rejected
and also 59049*3=1,77,147 (which exceeds the criteria of number being 3 digit)
So, n>999
and hence a bound between (1000,1000000)
"""
def digitsum(s: str) -> int:
"""
>>> all(digitsum(str(i)) == (1 if i == 1 else 0) for i in range(100))
True
"""
i = sum(pow(int(c), 5) for c in s)
return i if i == int(s) else 0
if __name__ == "__main__":
count = sum(digitsum(str(i)) for i in range(1000, 1000000))
print(count) # --> 443839